Question 1191710
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Derive the equation of the locus of a point P(x,y) which moves so that 
its distance from (2,3) is always equal to its distance from the line x+2=0.
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The line x+2 = 0 is the line  x= -2  (vertical line parallel to y-axis with x-coordinate of -2).


Let (x,y) be the point of the locus.  Then the distance from (x,y) to the line x= -2 is  |x+2|
(notice the absolute value sign).


The distance from (x,y) to the point (2,3)  is  {{{sqrt((x-2)^2 + (y-3)^2)}}}.


The equation of the locus is

   {{{sqrt((x-2)^2 + (y-3)^2)}}} = |x+2|.



Square both sides and get

    (x-2)^2 + (y-3)^2 = (x+2)^2

    x^2 - 4x + 4 + y^2 - 6y + 9 = x^2 + 4x + 4

    y^2 - 6y + 9 = 8x

    (y-3)^2 = 8x


It is final equation of the locus.  It represents a parabola with the horizontal axis y= 3, 
parallel to x-axis. The parabola is opened right. Its vertex is the point (x,y) = (0,3).
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Solved.